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Let $F_n$ be a free group on $n$ generators. Fix a prime $p$. Let $\gamma_k^p(F_n)$ be the mod $p$ lower central series, i.e. the inductively defined series $$\gamma_0^p(F_n) = F_n \quad \text{and} \quad \gamma_{k+1}^p(F_n) = (\gamma_{k}^p(F_n))^p [F_n,\gamma_k^p(F_n)].$$ Observe that the quotients $\gamma_{k}^p(F_n) / \gamma_{k+1}^p(F_n)$ are abelian $p$-groups. Moreover, the quotients $N_n^p := F_n / \gamma_{k}^p(F_n)$ are $p$-groups of nilpotency class $k$. They are universal with this property -- if $G$ is a $p$-group of nilpotency class $k$ and $g_1,\ldots,g_n \in G$, then there is a unique homomorphism $N_n^p \rightarrow G$ taking the generators of $N_n^p$ to the $g_i$.

Question : What are $H_2(N^p_n;\mathbb{Z})$ and $H_2(N^p_n;\mathbb{F}_p)$?

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  • $\begingroup$ Edited it because I realized that I asked about general $H_k$ when I really only care about $H_2$ (as indicated in the title). $\endgroup$
    – Tony
    Commented Mar 5, 2012 at 16:14
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    $\begingroup$ Did you try the Hopf Formula? Assuming you know the generators/relations. $\endgroup$
    – Chris Gerig
    Commented Mar 5, 2012 at 16:41
  • $\begingroup$ I tried. The Hopf formula seems to give me the answer for the quotients of the free group by the ordinary lower central series, but I couldn't make it work for the mod p version (for instance, the relations do not live in the commutator subgroup). $\endgroup$
    – Tony
    Commented Mar 5, 2012 at 21:17
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    $\begingroup$ I don't know the answer to your question, but I don't think that your comments about these groups being universal are quite correct. The quotients in your series are all elementary abelian $p$-groups. You need to define the $p$-nilpotency class of a group using central series with elementary abelian $p$-groups as factors. Your groups are then universal with respect to groups of $p$-nilpotency class $k$. $\endgroup$
    – Derek Holt
    Commented Mar 5, 2012 at 23:37
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    $\begingroup$ Dear @Tony: It might still be a good idea to add a correction to the question, as not everyone reads the comments. $\endgroup$
    – Ricardo Andrade
    Commented Sep 3, 2013 at 11:51

1 Answer 1

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For brevity, I denote by $F$ the free group on $n$ generators, $\lambda_k$ the $k$th terme of the $p$-lower series of $F$, and $N_k$ the relatively free group $F/\lambda_k$.

Also I use the following 5- terme exact sequence (see Generators of sections of free groups): $$0 \longrightarrow H_2(N_k;\mathbb{Z}/p) \longrightarrow \lambda_k/[F,\lambda_k]{\lambda_k}^p \longrightarrow H_1(F;\mathbb{Z}/p) \longrightarrow H_1(N_k;\mathbb{Z}/p) \longrightarrow 0.$$ Now since $N_k$ is minimally $n$-generated $p$-group, it follows from $ H_1(N_k;\mathbb{Z}/p) = N_k/[N_k,N_k]{N_k}^p$ that $ H_1(N_k;\mathbb{Z}/p) \cong (\mathbb{Z}/p)^n$, and also we have $H_1(F;\mathbb{Z}/p) = F/[F,F]F^p \cong (\mathbb{Z}/p)^n$. Now the above exact sequence reduces to $$0 \longrightarrow H_2(N_k;\mathbb{Z}/p) \longrightarrow \lambda_k/[F,\lambda_k]{\lambda_k}^p \longrightarrow 0$$

so we have $H_2(N_k;\mathbb{Z}/p) \cong \lambda_k/\lambda_{k+1}$, and this is completely determined by the rank $r_k$ of the elementary abalian $p$-group $\lambda_k/\lambda_{k+1}$.

The integer $r_k$ can be determined using Lie methods, see for instance Corollary 18 in "The automorphism group of a finite p-group is almost always a p-group" Journal of Algebra, Volume 312 (2007) G. T. Helleloid and U. Martin. In which it is proved that $$r_k = \sum_{i=1}^{k} 1/i (\sum_{j/i} \mu(i/j) n^j)$$ where $\mu(t)$ denotes the Mobius function.

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